If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is

Options

A. `4pi G//3gR`
B. `3pi R//4gG`
C. `3g//4pi RG`
D. `pi Rg//12G`

Answer

To find the mean density of the Earth in terms of the acceleration due to gravity (g) and the radius of the Earth (R), we can follow these steps:

### Step 1: Understand the relationship between gravity, mass, and radius
The acceleration due to gravity (g) at the surface of the Earth can be expressed using Newton's law of gravitation:
\[ g = \frac{GM}{R^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

### Step 2: Express the mass of the Earth in terms of density
The mass \( M \) of the Earth can also be expressed in terms of its volume and density:
\[ M = \text{Density} \times \text{Volume} \]
The volume \( V \) of a sphere (the Earth) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, we can write:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
where \( \rho \) is the mean density of the Earth.

### Step 3: Substitute the expression for mass into the gravity equation
Now, substituting the expression for \( M \) into the equation for \( g \):
\[ g = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} \]

### Step 4: Simplify the equation
This simplifies to:
\[ g = \frac{4}{3} \pi G \rho R \]
Now, we can rearrange this equation to solve for the mean density \( \rho \):
\[ \rho = \frac{3g}{4 \pi G R} \]

### Step 5: Final expression for mean density
Thus, the mean density of the Earth in terms of \( g \) and \( R \) is:
\[ \rho = \frac{3g}{4 \pi G R} \]